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Can a sound wave travel through a shockwave?
Say you have a scenario where there is a shockwave, and you sent a normal sound wave towards it. Could the sound wave travel through that shockwave? If so/not, why?
Would it depend on the strength of the shockwave? For example a sound wave might be able to travel through a weak shock but not a strong one.
(It also might depend on the strength of the sound wave itself)

I'm aware that shock waves themselves can pass through each other, (see the opposed wedges model here)
If that same thing applies to sound waves, that would indicate that they can pass through each other, but shock waves are a lot stronger than sound waves so I'm not sure this example would work.

Also, let's say for this example that the sound wave is approaching the shockwave from behind. In a real scenario that wouldn't be possible, but imagine somehow the sound wave could catch the shockwave and intersect it from behind.
I would also imagine some of the principles that apply to the sound wave intersecting the back of the shock would also apply for a sound wave intersecting the front.

Interesting if the sound wave would be refracted by the change in air density.

Towards it from the front, or from behind?  ;)

@MichaelHall Good question; it’d be interesting to see both scenarios. Probably from the back, as I’d imagine that would be the most common scenario on a plane (the engines making sound waves behind the plane shock)

I was joking.  Because a sound wave couldn't catch a shock wave from behind.

@MichaelHall Oh right, thanks for pointing that out haha. Still would be interesting to see (in a theoretical scenario) what would happen if the sound wave could catch the shockwave from behind.

"Still would be interesting to see (in a theoretical scenario) what would happen if the sound wave could catch the shockwave from behind.". Well, asking this indicates that you don't understand what a shock wave is. Can a boat's wake catch the boat ? Well, if the boat stops, or turns back into the wake, perhaps..  The shock wave is itself sound, (vibrations in the medium in which the shock wave exists). caused by the surfaces of successive expanding spheres of sound to pile up on each other.

The illustration that you use is actually less general than the question you ask. The illustration is of a situation where all the shocks are caused various part of the same flying object. I will deal with that case first.
Consider an observer moving with the shock surface and observing the flow in their immediate vicinity. They can measure the orientation of the shock surface, the flow vectors on either side, and all thermodynamic properties including the speed of sound waves. For them, the flow is steady. They will always observe certain facts. The ones that relate to your question are these.
The components of the flow velocity that are tangential to the shock surface will not change. The observer can also move with them, and the flow will now appear one-dimensional. It will be a "normal shock". The approaching flow will always be supersonic, that is, faster than the pre-shock sound sound speed. A sound wave can therefore approach from in front, and will be absorbed into the shock. The flow speed behind the shock will always be reduced, but may be supersonic (weak shock) or subsonic (strong shock), that is, faster or slower than the post-shock sound speed, which is itself always less than the pre-shock sound speed. In the weak shock case, no sound wave can approach from behind against the supersonic flow. In the strong-shock case a sound wave can approach from behind and is absorbed.
Summing up for shocks from a single object, sound waves can always approach shocks from ahead and are absorbed. The can approach strong shocks from behind and are also absorbed. Compressive waves strengthen the shock, expansion waves weaken it. The key is to realize that there is always an observer who sees the situation as one-dimensional.
More generally, consider two vehicles flying close together but not at the same speed or even in the same direction. There are now many more possible
cases. Broadly, the situation is that when one aircraft overtakes another,
their waves will momentarily coalesce, but if they pass in opposite directions, the left-pointing V will pass through the right-pointing V.
The whole thing is pretty-well understood, but you need a college-level text to give you the details. It might be more fun to spend an afternoon at your nearest duck pond. Wave behavior is pretty universal in all media, but surface wave speeds in a shallow pond are only a few feet per second!


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How does the downwash behind a swept wing differ from a normal wing?
How does the downwash behind a swept wing differ from a normal wing?
(I know you can’t trust everything you see on the internet, but this seemed legitimate)
Here it said:

The downwash pattern on a swept wing tends to increase the AoA towards the wingtip. We normally think of downwash as being caused by wingtip vortices, but really, we can visualize the entire wing as being constructed of an infinite number of vortices that get rolled up into one big one behind the airplane, so there is downwash created all along the wing. When you add the effect of all these vortices together, you get a different lift pattern along the wing than you do with a straight wing.

How does the downwash pattern differ for a swept wing? What along the wing does this?
Basically, why does a swept wing have different downwash amounts at different points, while a straight wing doesn’t?
Also, the same source said that this varying downwash makes the swept wing tip stall.

The only reason that I can think of is because the wing gets thinner as you get to the tip, so the downwash decreases.
Actually, this isn’t probably true. The downwash behind a wing is due to the wing making lift. Another way to put it is that the lift changes the downwash amount, not that the downwash amount changes the lift.

We often approximate a wing with a bunch of horseshoe vortices located at the quarter chord.  In the most simple approximation, we only use one such vortex.
Think of the trailing vortices as a whole bunch of vortex segments that lead off to infinity.  Each one is just a short segment.
That vortex segment has strongest influence perpendicular to itself.  Its influence drops off greatly before and after the segment.
When you apply these horseshoe vortices to a straight wing, the trailing vortices all start at the same location and all trail off together.  So, their influence on the wing is all very similar.
When you apply these vortices to a swept wing, the inboard trailing vortices start first.  They are in-line with the wingtips and have greater influence on the lift of the wing.  The vortices that trail from the wing tips start later and do not have the same alignment with the center of the wing in order to have an effect.
This causes a swept wing to have a substantially different lift distribution than a straight wing.
Usually a swept wing will be twisted to adjust the lift distribution to avoid any serious problems with stall etc.
Edit: Add image of swept wing horseshoe vortex system.

I see. So what effect does the bound vortex have? The bound vortex doesn’t actually exist if I’m correct. In other words, what flow changes direction because of the bound vortex?

(From what I remember it's just the circulation around the wing, but I could be wrong)

Everything effects everything.  Yes the bound vortex is a primary driver of the downwash, but the bound vortex does not have uniform strength.  The strength of the bound vortex is determined by the solution to the whole flow field.  (This is a case where my answer is leaning heavily on the mathematical abstractions we would use to model this wing -- it is a case where you solving a homework problem in a textbook would go a long way).

I think I understand it a bit more now. So if the trailing vortices of a horseshoe vortex system are caused by the circulation around the wing, why would there be a trailing vortex near the wing root? There is no tip circulation or voticity there, right?

Vorticity is shed anywhere lift is produced.  So there certainly is vorticity at the root.  I've added an image above -- the image isn't great, the 3D rotation provides a strange perspective.  Each horseshoe vortex has three legs, the bound leg and two trailing legs.  All three legs of a given vortex have the same strength.  Where there are two adjacent trailing vortices, they have opposite direction and they partially cancel -- which goes to show that the local wake strength is related to the change in circulation across the wing.

I see. So the bound vortex at the tip has less of an effect because the one at the center/root has the effect of effecting the ones around it? Like the tip has no bound vortex beside it, because it is at the tip. This would change the pressure distribution of the wing. Also, I'm assuming the pressure distribution would be so that the pressure would be higher on the upper surface towards the tip (compared to the root), and if that's the case wouldn't that help the tip stall later than the root? (Lower pressure on upper surface will help stalls form)

To make more sense, the bound vortex close to the center of the wing has the bound vortex beside it, and then that one has a bound vortex beside it, etc etc. These all influence each other. When you get to the tipward bound vortex, there is no bound vortex beside it, so it changes the pressure distribution.

No, Look at the region perpendicular to the tip bound vortex -- notice that it only includes other trailing vortices and the opposite tip.  Next, look at the region perpendicular to the bound vortex trailing from the root -- notice that it includes much of the wing as well as all the other trailing vortices.  The original question was about swept wings -- if the wing was not swept, there would not be this difference.

It is true that the adjacent trailing vortices almost cancel -- their strengths are almost equal, so the only part that remains is the difference.  However, the very tip of a wing produces zero lift, so the vorticity distribution trails off to zero there -- so the very tip vortex has very weak strength -- so the fact that it doesn't have a sibling trailing vortex to cancel with isn't super important.

Ah okay, I understand that more now. It makes sense how this would come together to change the lift distribution. Thanks.

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Does the back pressure on an airfoil stay the same with different speed?
(Inspired by this answer)
The back pressure is the force on the rear side of an airfoil (in the pressure recovery zone). When the air streams reach the pressure recovery area on an airfoil, they are met with a force, the back pressure. (this is due to the pressure recovery)
Basically it's the zone on an airfoil where the lower pressure, fast moving air meets with the surrounding atmospheric air conditions.
My question is will this force stay the same through different speeds? If not, why is that?
I see no reason for it to change, as the atmosphere will still provide the same amount of static back pressure at any speed, I think. I've seen here that the pressure recovery is also caused by the higher pressure under the wing coming into contact with the upper surface, lower pressure air. This might change things to where at higher AoA the pressure recovery gradient is steeper.

First, Cp changes along the airfoil.  So the pressure is different at different positions.
However, lets assume briefly that the Cp distribution stays the same with Mach  number (it does not).
When we dimensionalize Cp to Pressure
$C_p=\frac{P-P_\infty}{0.5\,\rho\,{V_\infty}^2}$
We see that for the same $C_P$ and freestream pressure $P_\infty$, then we will get a different pressure $P$ because $V_\infty$ is different.
So, when Mach number changes, the pressure changes.
In addition, the $C_p$ changes.
In aerodynamics, we can seldom assume things stay the same as other things vary.  When we can, it is usually a very clearly laid out principle that traces back to conservation of mass, energy, and momentum.

Ah okay. So do you mean the center of pressure when you say Cp?

Cp is the Pressure Coefficient @Wyatt https://en.wikipedia.org/wiki/Pressure_coefficient

@OrganicMarble Ah, I thought it might be something like that, thanks.

I think I get what you're saying. So because of the higher speed, pressure distribution changes, also changing the pressure distribution where the pressure recovery area on the wing is?

@Wyatt No, it is the change in Mach which affects the pressure distribution. As long as Mach effects are small, the distribution is much the same for different speeds. What does change is the pressure (as opposed to the pressure coefficient, which stays the same because the influence of speed has been removed by making it dimensionless).

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Does the flow turning amount change as you get further away from an object?
Will the flow turning amount change as you get further away from an object? This answer explains why shockwaves will extend past the body that made them (good starting point). This question is asking if those shocks will change flow turning angle as they extend outward.

Like this picture. The flow starts by turning at the tip of the triangle via a shock. Then that flow turns the flow above, and then the flow above that flow turns. Then Etc. Etc. Etc.
Say the angle of the triangle was 15 degrees at the nose. At the nose of that triangle, the flow would turn 15 degrees. At the back of the triangle, it also would turn 15 degrees.
The question is, would that 15 degree turning angle change as the shock got further and further from the object? If so, why?

Doing my research and looking through other answers, at the bottom of this answer it says:

The further we are from the nose, the smaller the turn of the airflow is and therefore the smaller the inclination of the shock wave is

I believe this to be true, but I don't know why this is the case. (That's what this question is trying to accomplish.)
Edit : After thinking about it for a bit, I came up with a theory. When the object turns the flow, it would be a hard surface doing it. When the shockwave turns the flow, the air is the thing turning the flow. Air is compressible at those speeds (Compressible in the sense of easy to 'squeeze') Because of this, the air might have an effect to where there is less turning.

This has entirely to do with whether the object is a 2D or 3D object.
The photo above might be 2D object, a long wedge viewed edge-on.  Or, a wedge between two plates of glass.
Or, the photo above might be a 3d object, a cone.
For the wedge, the flow turning continues to turn the flow, and the shock will be a straight line going off forever.
For a cone, there is a 3d relieving effect.  You can't just think about the flow in the plane that we're observing, but also what happens 'around' the cone.  As the distance from the axis of symmetry increases (i.e. the radius), the flow is allowed to occupy more volume.  Because of this, as you move out, the flow is required to turn less, which allows the shock to be weaker -- and so the shock bends.
Understanding this, we can see that the above photo is a cone -- not a wedge.
Most images you'll see online and in textbooks stop at the end of the cone.  It is important to look for images that continue past the end of the cone to make the best comparison.

I see. So basically because it’s a cone as you move out the flow has to occupy more surface area of the cone so it doesn’t turn as much?

That's about right.  Each 'strip' of flow really represents an annulus -- or a donut / ring.  The 'inner' rings have much less area than the 'outer' rings.
For wedge flow, each strip is a strip and is identical to every other strip.
Supersonic cone flow is very different from supersonic wedge flow because of this.
In general, this idea is called '3D relieving' and it influences lots of aerodynamics.  Imagine a NACA 0012 airfoil and another, but revolved into a 3D streamlined body.  The Velocity peaks around the airfoil will be greater than the axisymmetric body.

That's interesting. I had thought of that idea before, but I never really determined that the air would 'spread out'. Also interesting to see how that affects shockwaves and other phenomena.

This is very easy to see in cylindrical coordinates.
In normal (Cartesian) coordinates, the volume of a chunk of flow is dV=dxdydz.
In cylindrical coordinates, the volume of a chunk of flow is dV=dxrdr*dTheta.
The difference is that 'r' term.  As 'r' increases, the volume in an otherwise similar 'chunk' of space gets bigger.
Think of it like flow in a pipe, but now the pipe expands like a funnel.  If water were flowing through the pipe (constant mass flow), it would slow down as the funnel gets bigger.
Flow in a variable area pipe is called 'Quasi-1D' flow.


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Why doesn't more resistance lower on a plane (or higher), for example an airbrake sticking way down, make the plane have a pitch down moment?
What initially sparked my question is when I saw the Redbull plane swap stunt, where there was an airbrake deployed on the bottom of the plane. Why wouldn't this cause there to be more drag lower down and cause a pitching moment? Also, why does moving the CoP longitudinally increase and decrease stability, while moving it along the vertical axis doesn't?
Also, say you have an external fuel tank or an airbrake, like I mentioned. When pulling up, there would be increased flow over it because it's further away from the CG, which would cause more resistance, correct?
Lastly, what about planes that have a high wing config, does the resistance from that create a pitching moment?

Why do you think it doesn't have a pitching down moment? Most likely it does and pilot just trimmed it out.

@Chris oh okay, what about planes that have a high wing config, does the resistance from that create a pitching moment?

Any force with a component along the longitudinal axis that acts above or below the center of gravity has an associated pitching moment. In unaccelerated flight, all of these pitching moments are in equilibrium.

@Chris Oh okay, makes sense. "In unaccelerated flight, all of these pitching moments are in equilibrium." I'm assuming that's after the pilot has trimmed everything correctly, right?

Yes, that's right.

@Chris alright thanks for your help!

So as your intuition has clearly told you, there will be a change in pitching moment, you're correct, however, exactly how big this change is depends on the aircraft and how it's being flown. Im not familiar with the Red Bull stunt, or the aircraft involved, but if the airbrake is sufficiently near the centre of pressure (CoP) on the airframe, even if it, say, increase drag by 30%, if the airbrake is only 20cm from the CoP, the CoP won't move very much, and the pitching moment will be quite small. Equally, as some commenters have pointed out the stunt pilots will have practiced this many times, and may know precisely how much back stick to input to counter the airbrake...
Now onto the way CoP position affects stability. you may already know, but if an aircraft's CoP is behind it's Centre of Mass (CoM), its is considered stable, this is because if there is a sudden disturbance (say a gust of wind) causing an increase in lift, the aircraft will pitch down, reducing the angle of attack, and reducing the lift produced, countering the effect of the disturbance, likewise if the CoP is in front of the CoM, a disturbance will cause an exacerbating motion, requiring sophisticated active controls (many computers) to counter. (This is a slight over-simplification, obviously, in level flight the CoP is exactly on top of the CoM -or else there would be a moment on the aircraft- since the true CoP is the sum of all pressure forces, wings, tail, the lot, so in reality we are talking about the positioning of the main aerodynamic devices on the aircraft, in particular the wings, in relation to the CoM). Moving the CoP up or down will not have any significant effect on stability because the lift force acts perpendicular to the direction of flight, so the moments involved are only interested in the perpendicular distance from the CoP to the CoM, which in this case is longitudinally. Now, there is also the drag force, which will produce moments depending on where the CoP is up or down, since it acts parallel to the flow, but it tends to be an order of magnitude smaller in size than the lift force (typical airliner Lift to Drag ratios are about 17-18 in cruise, gliders closer to 20 and fighters more like 12). In all off this, remember that aircraft have tail surfaces though, which are constantly being moved about to tweak the CoP and put it precisely in the place the pilot requires, either by direct use of control surfaces or trim, depending on the size of moment the pilot wishes to produce or counter.
In terms of increased 'flow' when pulling up, I'm not quite sure I totally understand. It is true that deflecting a surface to produce a moment will turn the aircraft in relation to the airflow, and this will change the area exposed to the airflow, or increase the drag coefficient, or both, this will change the size of the drag, and also the size of the moment on the aircraft, but again, most forces don't produce huge moments on the aircraft, and if they so, there are always big control surfaces to counter. This is perhaps obvious, or aircraft would depart from flight all the time!
I hope some of that is helpful!


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How do transonic or supersonic speeds affect lift distribution of the entire wing?
I've written a few questions about this topic, but this part I never asked.
How do supersonic or transonic speeds affect lift distribution over the whole wing? How does the root's changing pressure distribution compare to that at the tip?
In other words, why do supersonic or transonic speeds affect the wing differently over span?
(This might seem like a duplicate of my other question, but it's asking something different)

In pure subsonic flow the pressure distribution is the combination of the flow around a body, where air has to speed up locally to make way for that body, and the flow around an inclined flat plate. Things will become more complicated if the rear end of the body is blunt and causes the flow to separate locally.
If you raise the Mach number, density changes will become greater and will require larger speed changes to let air flow around the body. Please remember the tube explanation from this answer to explain what happens in detail.
Things become really different once the local flow speed becomes supersonic. When before a contraction in the body would cause the flow to slow down, it will now accelerate further. This will lead to a local drop in pressure when before, at subsonic speed, pressure would increase. Eventually, this supersonic pocket will collapse in a shock, so speed drops to subsonic while pressure rises immediately. Below I copied an example of typical speed distributions from US patent 3,952,971 by Richard Whitcomb.

Speed and pressure distributions are very similar, only that pressure changes with the square of speed.
This extension of the low pressure area on the wing will shift the center of pressure backwards. If you increase flight speed further, the supersonic area will increase until at Mach 1 it will extend over the whole length of the airfoil.
On a straight wing this will happen the same way over wingspan. Only on swept wings will sweep effects make root and tip pressure distribution look different. But what happens can all be explained by the same basic principles:

Low pressure areas suck in more air, so they attract the flow.
Conversely, high pressure areas will divert the flow.
At subsonic speed air needs to speed up to make way for the airplane.
At supersonic speed air needs to slow down to make way for the airplane.


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Could the radar on a passenger plane be used as a makeshift AWACS?
I'm aware that an airliner doesn't have a way to transmit data over datalink, but could an airliner at least tell other aircraft what they see on radar? Is an airliners radar powerful enough?

(Source : Link)

Many aircraft do have a ‘datalink’ of some kind, but looking at the radar screen and describing by radio what the pilot sees is quite common.  The type of radar used on AWACS is far higher spec than that found on an airliner though, so it really depends on what ranges and level of detail is needed in any particular scenario.

@Frog Yeah I was thinking if a friendly airliner went close to a combat zone, it could relay target locations to friendly fighter jets.

I don’t know for certain but I’d be confident that fighters have at least as well-suited radars, and would be more willing to fly in/near a combat zone

@Frog haha definitely, but I was just curious if it was possible.

That's a weather radar. Used to detect weather, not aircraft.

In short, a weather radar is not able to detect small targets. Clouds are large volume objects, not comparable in size with an aircraft. It is very unlikely a weather radar could detect another aircraft, even at a distance of 2 NM.
(Let's put aside an AWACS is a command center and a tactical communication hub coupled with a multimode radar.)

Source.
Radar cross section vs. beam size
A weather radar scans the sky with a large beam, 4° to 10°. A cloud can intercept the totality of the beam energy. In contrary, an aircraft can only intercept a tiny fraction of this energy. This quantity when back-scattered is likely below the noise level of the radar receiver.
Radar frequency and beam size are selected in relation with the resonant frequency and the size of the target equivalent dipole. Weather radars are designed for water resonance and large packets of water droplets. They are not designed for targets of small radar cross section (RCS), e.g. a fighter with a RCS of 5m$^2$.
This is the main obstacle for aircraft detection, but there are also other points to be considered.
Range and resolution
If we take the E-3, the main radar has a 9m rotating passive antenna array working either in C band (around 5 GHz) or S band (3 GHz), which are relatively low frequencies, selected because they allow more powerful transmitters and the signal is not significantly attenuated by the atmosphere. This is required to detect targets at more than 200 NM, which can be stratospheric missiles, low-flying aircraft and ships. The higher power requires a large antenna, this size in turn improves the angular resolution, even more in C band.
Weather radars work in S band, but with their small antenna, and their limited power, they don't have the previous advantages.
Angular coverage
The horizontal scan of a weather radar is about 120°, 1/3 of an AWACS. In the vertical direction, a weather radar has a very small aperture, limiting detection in a small vertical range and certainly not ballistic missiles or aircraft at the forefront of the ground surface.
Doppler detection
Clouds appear moving at the velocity of the aircraft. Most weather radars take advantage of this property, and use a Doppler shift measurement to detect clouds and rainfall.
A target moving parallel to the plane, would have a null Doppler shift and would be rejected by the radar processing unit. This could be used by the target to escape detection.

The pilot can tell other aircraft what he sees on the radar. However, the weather radar is only really designed to detect dangerous weather.
There are multiple functions an airliner's weather radar lacks for Airborne Warning (AEW) use:

No ability to identify and IFF-interrogate other aircraft.
No tracking mode on the radar. Military radars first scan for targets, then switch to tracking them. (Advanced radars can also track while scanning).
Less ability to ignore weather due to a wider aperture.

We're talking AW, the first two letter of AWACS. AEW&C aircraft also carry staff with workstations to control and manage traffic. Datalinks are actually optional; they can transmit target coordinates in a way that allows a compatible radar to immediately track it without scanning. But voice communications are generally enough and more important.
In a practical scenario, a fighter radar will most likely be able to detect targets from a longer range than the airliner + its direct communications range (without relaying). One of the few exceptions I can think of is if an airliner is being escorted by a stealth fighter, and relays that information.

No.
Much, much more than a similar airframe is needed.  Consider that:

That giant Frisbee on the top has 360 coverage instead of a small cone out the front.
It's way more powerful.
There are all kinds of electronic equipment installed for signal processing, target interrogation and identification, plus encrypted voice and data communication, most of it classified as Secret or Top Secret.
There is an entire crew trained in the use of the equipment listed in #3, not just a couple pilots keying the mic and saying "hey, I got something out there..."

The AWACS is a fully engineered and integrated military mission system that can't be replaced by a simple weather radar in the nose cone and a VHF com radio.


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Is there any video of an air-to-air missile shooting down an aircraft?
I was curious, is there is any video of a missile shooting down an aircraft?
I have looked around, and I can’t seem to find anything substantial.
(Anything I did find was fake, such as this)
Air-to-air is what I’m looking for, but surface-to-air also counts.

@Wyatt I'm not sure of the two downvoters, but I think this is a bit off-topic (only a little). Also, it lacks research effort, there's quite a lot of actual footage if you search around the web. Finally, it's too broad. All the more reason to downvote, since it is the combination of all downvote reasons.

@AircraftEnthusiast007 If you hover over the downvote button, it says to downvote if there is no research effort. I did research, evidenced by the linked video of my failed attempt to find what I was looking for. This question isn't really something I could've made less broad. I agree that it's a bit short, but I honestly don't know what else to add (suggestions welcome). Also, there are plenty of other missile-related questions on here if you search around, so I don't think it's exactly off topic by that measure.

Unfortunately, missile related questions are often downvoted without providing a reason. However I would like to point out that not only are missiles accepted to be aircraft (per this question answer) but theses types of questions are also voted to be on topic, per this meta question

I am a bit uncertain what exactly you are searching for, however you can find  plenty of videos on the internet which show missiles hitting their target.

The AIM-9X Sidewinder development test video show the intercept of a F4 Phantom II by a Sidewinder AIM 9X infrared missile. Additionally, the infrared image of the missile is shown, which was telemetered down as part of the test suite, which is very interesting to see as it simultaneously gives the perspective of the missile. In my opinion, this is the best unclassified video showing an air-to-air intercept available.
The Russo-Ukrainian war is probably the most documented war ever. This also includes missile intercepts such as this video depicting the downing of a Mil MI-24 Helicopter
And of course a lot of videos exist of the Iron Dome system, which produces a big firework when intercepting incoming projectiles.

This is exactly what I was looking for. The question might’ve seemed broad, but this is what I was looking for. Thanks

One place you can find a lot of classic air combat footage is in the old Discovery Channel (back when it was good) documentary show Wings. Pick an episode about a missile-armed aircraft and enjoy an hour of aircraft and combat footage, plus 80s TV commercials! I picked the famously missile-only F-4 Phantom II and a quick skim caught footage of multiple air-to-air kills at 22:10. Unfortunately, most of the footage is unattributed so you rarely know what you're actually looking at.
Wings Playlist on YouTube.

Interesting footage. I honestly was trying to compare the ‘shoot down’ physics of games like DCS to the real world, and it seems fairly accurate!


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Why will a positive stagger increase biplane efficiency, while a negative stagger will decrease efficiency?
On a biplane, why will a positive stagger (top wing more forward relative to the bottom one) be more efficient than a negative stagger? In both cases the wing’s pressure fields are more separate, so why does it matter if it’s positive or negative stagger?
In other words, why will a negative stagger decrease efficiency while a positive stagger will increase efficiency?
What is the difference?

Note: I have read the answers to this question, and this question actually was sparked by one of those answers. It explained that positive stagger will increase efficiency and that negative stagger will decrease efficiency, but it didn’t explain why, which is what this question is asking.

Does this answer your question? Why do most biplanes have their top wing slightly forward of the lower wing?

@sophit: Actually, I think this question derives from one of the answers to that question. It uses the same image. The linked PDF makes the claim that negative stagger reduces lift and efficiency without explanation.

@FredLarson That is exactly correct.

@sophit FredLarsons assumption is correct. The answer(s) you linked actually sparked this question.

(Also if anyone has a vote to close and you see why this isn’t a duplicate, I’d appreciate it to be removed so this doesn’t get closed, thanks)

While not an answer, you can just look at the picture and see some of the answer. Wings deflect air above them more than below them. The bottom wing will influence the air stream of the top wing. Moving the top wing forward compared to the bottom wing will put the top wing in "clean" air. Moving it backwards puts it further into the disruption caused by the bottom wing. This may help: https://aviation.stackexchange.com/questions/27441/how-far-away-from-a-wings-surface-does-the-air-stream-remain-unaffected

With negative stagger, perhaps the lower wing makes a shadow on air flow to wing above, reducing efficiency?
A bit as 'Super Stall' in F-104 and other T tail machines. A wind tunnel test could provide clues.
Blessings +

With advanced lower wing, it can make 'Shadow' on upper wing, reducing efficiency, more possible as AoA increases. A bit as Super Stall in F-104 and other T-tail machines. To be confirmed in wind tunnel. Or not?

Because we look only at efficiency at positive lift.
Remember that increasing the distance between the wings of a biplane increases efficiency (see for example Table III in Max Munk's NACA report 151). And now consider what the air "sees" as it approaches the staggered biplane at a positive angle: Positive stagger will make the distance between the wings appear larger and vice versa.
To illustrate my point I have made 3D of two wings in Tinkercad:

First the side view. Span is 8.6 meters, distance is 1 meter, stagger is also 1 meter.

Now the view from the front at a positive angle.

and lastly the same at a negative angle. Notice how the two wings seem to move closer together.
That is all! Aerodynamically, the distance, and, consequently, the interference, between the wings changes with the sine of the angle of attack and proportional to stagger.

I see, thanks for your answer. So is David's comment on my question also correct? In that the upper surface deflects more air, so it has more of an influence on the air above than below?

Brilliantly simple!  What does the oncoming air see.

@Wyatt No, I'm afraid that "top defelcts more than bottom" theory is wrong. I think interference is lower with positive stagger and positive angle of attack. If you look at efficiency at negative lift coefficients, negative stagger would show better efficiency.

This answer isn't an intuitive explanation of why, just a set of numerical observations that support the claim.
NACA Technical Note no. 70, "The effect of staggering a biplane," by F.H. Norton, 1921.

This investigation was carried out at the MIT wind tunnel in 1918.  ... The lift, drag, and center of pressure travel are determined for a biplane with a stagger varying from +100% to -100%. It is found that the efficiency and the maximum lift increase with positive stagger.

Fig. 4 plots L/D vs. stagger for AoAs from -2 to 18 degrees.  At moderate stagger (-50% vs. +50%), for AoAs from 0 to 10 degrees, L/D improved by 20% to 1%.
(Fig. 2 plots CL vs. stagger, ignoring CD and thus ignoring efficiency.  Especially at AoAs exceeding 10 degrees, CL increases greatly with positive stagger.  This trend may inspire an intuitive explanation.)

Unfortunately this does not answer the question which is why is positive stagger more efficient.

Yes, as stated.  But it might inspire a more targeted answer, and it's too long for a comment.

I've become skittish about "why" questions (elementary "why do foils cause lift" or trickier ones about tip vortices and sweep and downwash) whose answers can validly range from "just navier-stokes it and see" to utter albeit inspiring handwavium.

Still, and answers should at least try to match the question. The document you linked is easily found, yet I decided not to use it as an answer because it does not provide answer to this question. Why staggering does what it does is a very valid question despite its simplicity.


text:
What causes the vapor cone when an aircraft is traveling at M ≤ 1?
What causes the vapor cone of an aircraft going M ≤ 1? I know that the vapor is from the air expanding, but why is it in the cone shape? What shock system forms it into that shape?

There is no vapor cone at Mach > 1. Condensation cones (for airplanes) or condensation collars (for rockets) only happen at transsonic speed, a speed near but still below Mach = 1.

Your original version was correct.  This is a subsonic aircraft.  M<1.

@RobMcDonald oh right, fixed.

You should check out this reference.  It goes into great detail of exactly what is going on in photos like this.
The flow is accelerating as it goes around the aircraft.  As it accelerates, the pressure drops and the vapor cloud can form.  In this case, the flow accelerates to supersonic speed and then is shocked back down.  The shock forms an abrupt pressure rise which causes the termination of the vapor cloud.
The flow does not have to reach supersonic flow or have a local shock for you to see a cloud like this -- but the abrupt termination of this cloud certainly looks like this one is terminated by a shock.


text:
Why is a half-streamlined body less aerodynamic?

As seen here, a streamlined half body is less aerodynamic than a fully streamlined body. The half streamlined one has less surface area (kind of) and less frontal area. Why is its drag more?

What is the source of this graphic, and what do the hash marks under the streamlined half-body (which isn't the term used in the question title, btw) represent?

These are coefficients. What is the reference area? How do you know that the last body in the list has less frontal area than the one before?

@RalphJ Honestly not sure, but I got it from here

@PeterKämpf Yeah good points. I guess I just assumed that they considered that, but maybe not. Got it from here

@quietflyer thanks for pointing that out, fixing it now.

@RalphJ - the hash marks represent a wall.  The part of the half-body that is touching the wall isn't exposed to the airflow.

If it wasn't against a wall, what should the number be instead of 0.09?

The drag coefficients are normalized by a reference area.  For something like this, it is going to be the frontal area.
So, if the streamlined body and half-body had the same frontal area, the drag on the half body would be larger.  That isn't particularly surprising.
This figure is misleading because all the objects are not drawn to equal frontal areas.
It is hard to know what differences there are in the half-body shape -- for example, does the aft end close out in a 2D or 3D manner?  I.e. is it half of an axissymmetric body, or more like a 2D airfoil shape with the nose rounded over.
Edits:
Here is a quick model of a streamlined body and matching half-body I threw together.

These two bodies have identical frontal area.  Therefore, their drag coefficient ratios should equal their drag ratios.
First, to achieve the same frontal area, the linear dimensions of the half body were scaled up by $\sqrt{2}$ (i.e. 1.414).  So it is 41% longer and 41% wider than the full body.
The wetted area divided by the reference area is the same for both bodies (the flat area against the wall is not counted against the half body).
$S_{wet}/S_{ref}$ = 10.689
So, these two bodies have identical wetted area.
If we assume the wall extends far from the body, it will act like a symmetry plane -- and the inviscid portions of the flow (or the entire low if it was inviscid) would be identical.  I.e. we expect the pressure distributions around the two bodies to be similar.
This leaves us with differences in Reynolds number.  At the same flow condition, the half body will have a Reynolds number 41% higher than the full body -- which could place it in a dramatically different place.
Of course, from the original drawing, it is possible the half body looks like this:

Notice the 'flattened' tail.  You can see it from the top view, but the side view looks identical.

I removed the confusing sentence about equal scale bodies and instead added figures illustrating the difference in size for equal cross sectional area bodies.

I'd bet money that the streamlined half-body is meant to be as in your first case-- it's the simplest interpretation. Still, we don't know for sure do we

I agree, it is likely as the first case.  However, I bet the second case has lower drag.  It has more wetted area -- but, it closes out the aft end more gently and will therefore be less likely to have separated flow.  The point of including it however is that we don't know because the figure is not clear.

The numbers given in the table are coefficients, not force values.  They have to be referenced to a given frontal cross-section area.
The wrong way to interpret the table:  "If I take a streamlined body, and then I cut it in half and plop it down against the floor of a wind tunnel, then the drag will have increased by a factor of .09/.04 = 2.25."
The right way to interpret the table: "If I take a streamlined body, and then I cut it in half and plop it down against the floor of a wind tunnel, and then I magically expand the size (volume) of the body so that its new cross-sectional frontal area is exactly the same as before I cut it in half, then the drag will have increased by a factor of .09/.04 = 2.25."
Here's an observation: using an ellipsoid (for simplicity) rather than a streamlined body, when we go through this procedure, the actual exposed surface area remains exactly the same, not counting the surface area in contact with the floor.
For example, an ellipsoid with a major axis of 3 units, and with minor axes of 1 unit each, has a surface area of 30.83 units, and a frontal cross-sectional area of pi.
If we want to enlarge the ellipsoid so that (before slicing in half) it has double the frontal cross-sectional area, the minor axes will be enlarged to each be the (square root of 2) units long, and the major axis will be three times that length.  The surface area is now exactly doubled, to 61.66 units, so after cutting it in half, we are back to a surface area of 30.83 units, not counting the surface area in contact with the floor.1
However, we've increased the enclosed volume by a factor of 1.414, or the square root of 2.
These relationships remain true regardless of the ratio between the length of the major and minor axes of the ellipsoid.  For example, if we increase the initial length of the major axis from 3 to 5, while leaving the minor axes at length = 1, then after we expand the ellipsoid to double its cross sectional frontal area, and then slice it in half, we'll still end up with the same surface area we started with (not counting the flat side), and we'll still end up with 1.414 times the enclosed volume that we started with.
The surface areas and volumes noted here were calculated by this on-line calculator, which also states the formulae involved.
Since we increase the enclosed volume when we expand the streamlined body to double its cross-sectional frontal area and then slice the object in half, it's not surprising that we'll find that the drag coefficient has increased.  There's more "bulk" there to deflect the streamlines.
Footnotes:

Another related answer has noted that this same relationship-- total surface area remaining constant, not counting the area in contact with the floor-- also applies when we apply the same manipulation to an actual streamlined body, rather than to an ellipsoid.

I'm reconsidering this answer.  So there's more volume enclosed by the half-body than the full body, for the same front cross-sectional area.  Maybe that should just provide a more efficient way to fair the shape and reduce drag.  If the front cross-sectional area is the same in both cases, and the exposed surface area is the same in both cases, why should there automatically be a higher drag coefficient, and more drag, in the half-body case?  I'm not sure this answer has as much explanatory power as I thought it did when I created it.

Maybe a better perspective is that, for the same cross-sectional area, a circular cross-section is a more "compact" shape than a semi-circle, and so the circular cross-section is likely to be able to move through the air with less disturbance, if other things like the taper ratio of the shape, the total exposed surface area, etc are all equal.

In terms of drag coefficient, it is important to have "clean", undisturbed air flowing over a surface to minimize drag.
The half airfoil shape near the ground has a higher coefficient of drag because airflow near the ground is more chaotic than airflow further away.
Through no fault of the shape itself, the measured drag coefficient will be higher near the ground.
Move the same shape away from the ground, maintaining 0 angle of attack, it's drag coefficient will be lower, especially if the bottom leading edge is rounded.
One can see why engineers of the 1940s became enamored with "laminar flow" to reduce drag for long distance aircraft.  Turbulence disrupting the boundary layer can dramatically increase drag.
Scale can also affect drag coefficient because, again, of greater distance from the ground.  The 1938 Volkswagen Beetle has a drag coefficient of 0.48.  The more modern Volkswagen XL1 has a drag coefficient of 0.19, and the Ecorunner 8 has 0.045.
Finally we can see the number values in the question chart refer to coefficient of drag, seen in the drag formula as:

Drag = Cdrag × Area × Air density × Velocity$^2$

To get a comparison of Drag for given shapes, the frontal areas must be equal.

Re "The half airfoil shape near the ground has a higher coefficient of drag because airflow near the ground is more chaotic than airflow further away."  -- in a wind tunnel shouldn't the airflow at the wall be laminar?

@quietflyer generally, laminar flow will exist only for a limited distance from the leading edge.  Friction interaction of a fluid and a surface is also a concern in pipes.

For a symmetric streamlined body in the middle of the wind tunnel, the streamlines are presumably symmetric around a plane of symmetry and don't cross that plane at all. Imagine that we turned that plane of symmetry into a very thin wall. One might naively assume that the streamlines would continue to flow as before and total drag would be the same, that is, we would have two half-bodies each with half the frontal area each contributing half the original drag. But in fact we get 2.25 times the original drag from the two sides combined. So the wall must do something dramatic to the airflow.

This is a typical example where a picture is worth a thousand words... or is it?
This picture is a potpourri of several experiments, some of them performed and recorded in very old NACA reports. The coefficients for the circular shapes have been extracted for example from this NACA report dated 1937.
The coefficients in that picture are valid for fully turbulent flow, at Reynolds numbers from $10^4$ to $10^6$ and, as usual for subsonic aerodynamic coefficients, they have been adimensionalised via the "frontal area". This explains the difference between the "streamlined body" and the "streamlined half-body": the latter possesses half of the frontal area and therefore its coefficient would be twice as much as the full body, even if they had the same drag.
Even if it's not specified, the coefficient of the cube has been obtained from a cube lying on the wind tunnel's wall.


text:
Why are the top of the F/A-18 strakes curved?

As seen in this picture, the top of the strake has a slight curve to it. I'm mainly talking about the curve seen if you were to look at it straight on.
To be more clear, not the curve of the strake as you travel down the airframe, but the downward curve if looked at from the side or front. Basically where the strake attaches to the airframe is at a higher point than where it ends. (Not the best picture for this, but it's the best I could find)
Is there any aerodynamic reason for this, or is it just how the airframe was built?
If I'm correct, the curve would raise the pressure on top of the strake, as a curved surface would be closer to the freestream air.

Wyatt, please do not continue to make small edits to your question, bumping it to the top of the "active questions" list with each one. That's considered poor form and can have adverse consequences.

@RalphJ Sorry about that, probably should've edited everything at once.

The airflow at some point over the top of a wing that is close to a fuselage will not necessarily be in the exact direction of flight. You would then expect the strake orientation to be finessed to follow the airflow, so as to minimize yawing forces and drag.
Now note that strakes are commonly used to correct the direction of airflow over the wing, so as to minimize turbulence propagation into the airflow over the tail surfaces at large angles of attack. God and the computational fluid dynamicists only know what the spanwise airflow distribution is under those conditions, and whether or not the strake requires being set at an angle to the direction of flight to ensure its effectiveness.
As always, Peter Kämpf knows the right answer and your mileage will be lower in California.


text:
Which is more efficient, thickening an airfoil or adding a bottom-surface curve?
So if you were to take a standard airfoil and increase the thickness, you’d increase the lift (Bernoulli’s principle), but it would also increase the drag.
Now imagine you have the same airfoil, but instead add a curve to the bottom instead of making it thicker. Doing that would increase the lift also (Newtons 3rd law).
So to achieve say 5 pounds of lift extra, would it be more efficient (less drag) to make the wing thicker or add a curve to the bottom, similar to the curves seen in supercritical airfoils?
Edit : What is meant by “add a curve to the bottom”

Not a bad question really.  Much of airfoil design and performance depends on airspeed.  Generally, if one wants more lift go faster, as lift is proportional to V$^2$.  Wing design depends strongly on Reynolds number.

So if you were to take a standard airfoil and increase the thickness, you’d increase the lift

No, you don't. Lift is determined by camber and angle of attack. Both don't change with thickness, so lift also does not change. However, the thicker wing allows for more spar height, so a wing of the same area can be built lighter. This gives you less wing weight and more lift remains for carrying fuselage, fuel and payload.
Drag will also rise, dramatically so at transsonic and supersonic speed.

Now imagine you have the same airfoil, but instead add a curve to the bottom instead of making it thicker. Doing that would increase the lift also

Wrong again. Now you decrease camber, so the wing will create less lift at the same angle of attack and dynamic pressure. All what I said above also applies to this case.

New comment after the editing of the question:
The airfoil you show is by no means a standard wing airfoil. It looks more at home in a gas turbine. Such very highly cambered airfoils are not used in airplane wings precisely because they only work well in a very narrow angle of attack range. Generally, filling up the hollow bottom will widen that range and reduce drag at lower angles of attack. It also will make them look more like a standard airfoil (that would be older ones like the Gö 549 and the Clark Y, both of which were used to define the 4-digit NACA range). Still, it will result in a much too thick airfoil and will show separation over the full angle of attack range due to its excessive thickness. Drag will be horribly high.

Oh okay, thanks. So I always thought adding a curved bottom to an airfoil would increase lift. Isn’t that how supercritical airfoils compensate for the flatter top?

Also isn’t adding a curve to a supercritical airfoil what makes it have high rear loading? (See new edit to know what is meant by “adding a curve”)

@Wyatt Don't add or subtract curves, but define a camber line and add an aerodynamic shape around it.

Ah okay, yeah that edit made it more clear. So basically a higher cambered airfoil (like the one in the picture I edited in) will make more lift, but only at certain AoA ranges? (Last question because I don't want to keep asking tons of questions)

@Wyatt Yes, right.

Oh okay, thanks!

So if you were to take a standard airfoil and increase the thickness, you’d increase the lift.

This is true only till around 12% thickness, afterward the $C_{l_{max}}$ decreases with increasing thickness. This plot shows this behaviour quite clearly (source):

Now imagine you have the same airfoil, but instead add a curve to the bottom instead

This would also increase thickness unless the shape of the bottom is concave (or unless you mean something else).

So to achieve say 5 pounds of lift extra, would it be more efficient (less drag) to make the wing thicker or add a curve to the bottom, similar to the curves seen in supercritical airfoils?

Increasing the thickness increases drag, mainly because the airfoil has a bigger surface and a bigger pressure recovery on the rear part (and a bigger divergence drag if used at transonic speeds). So if you can choose then it's better to trim the camber of the airfoil than increase its thickness.

Note, this is a chart for CLmax, not for CL at fixed alpha.  Perhaps that is what the OP was asking about, but the question was far from clear.

@RobMcDonald I added a picture to maybe clear some confusion up.

No, your figure adds nothing to the conversation.  The curve of an airfoil is defined by the mean camber line.  The thickness is measured from that.  You don't just add thickness or curvature to the bottom or top.  Thickness is added to the camber line - which is the middle and can be curved.

The other potential confusion is whether you were asking about increasing lift at a fixed angle of attack -- say how much lift at three degrees.  Or if you were asking about increasing the maximum lift possible that can be attained by a particular airfoil at any angle of attack.  Peter and I are answering about lift at a fixed angle of attack.  sophit is answering about maximum lift capability.

I think @sophit is referring to coefficient of lift, which is fairly linear in the useful AoA range.

@RobMcDonald: both comparisons ($C_l$ Vs $\alpha$ and $C_{l{max}}$) are tricky. In my opinion the plot I posted makes anyway sense since the comparison is done within the same "family" of airfoils.

@sophit I am totally fine with your answer -- it is probably better than mine (if that is what he is asking).  Thickness influences CLmax, but not CL at a given alpha.

Adding thickness doesn't add lift.
Don't cite Bernoulli's principle and Newton's 3rd law as causal in an aerodynamics discussion -- everything is far more complex and nuanced than that.
You can't add curvature to just the bottom.
You should learn about thin airfoil theory.

You were doing fine until you threw in "supercritical airfoils".  Now we have to go over all 130 years of modern aviation.  A bit much for one question but ...
Yes, adding undercamber will increase the lift coefficient.  This is what we do with slats and flaps.  Supercritical airfoils add a little undercamber in the back to help balance the center of pressure with the lift created by the top front portion of the wing.  By design, most of the upper curvature, or camber, must be removed from the top of the wing to avoid a draggy supersonic shock wave from forming there.
Going back in time to the Davis airfoil, flying at less than Mach 0.5, designers took full advantage of upper curvature because it by far has a higher Lift/Drag ratio than lift produced by undercamber as long as the Reynolds number is high enough (> 500,000) to use it.
Designers of the Davis airfoil found that moving the area of maximum camber back a little delayed onset of drag producing turbulent flow (increasing span and shortening chord (higher Aspect Ratio) does the same).
"Laminar" wings, essentially turning a classic airfoil backwards, were tried.  The stall characteristics proved difficult to manage but the idea can be seen in ... the hull of an Iowa class battleship!  A sharp leading edge with very little camber will make the flow of the working fluid smoother along the surface, minimizing disruption of the boundary layer.
Further back in time, when airspeed was not high enough to take advantage of "top lift", undercambered wings were favored.  Not surprisingly, early aviators modeled their wings based on birds.
In the second decade of the 1900's, designers found that thicker heavily cambered wings with convex bottoms were not only stronger ... but also had lower lift to drag ratios! (if they went fast enough).
Amazingly, some held on to the belief that thinner wings were better, and, when Mach effects became significant, they had their day.


text:
Is air that is accelerating more turbulent?
Does air that is accelerating have more turbulence? It makes sense why air decelerating would make turbulent flow (e.x, the air during pressure recovery over the rear of an airfoil). Would speeding it up make it turbulent? Also, does the air being in a confined space (for example a tube with airflow) have any effect on this? Compared to unconfined airflow, of course.
(All of this assuming there is no object obstructing flow)

No.
Turbulence means that air has significant speed components in directions orthogonal to its primary direction of movement. A negative pressure gradient accelerates air, and it does so only in the direction of that pressure drop. This means that the whole air picks up speed in one direction only and the speed components orthogonal to this direction become smaller relative to the main speed component. This reduces turbulence.

As an example, one way that a wind tunnel reduces (undesirable) turbulence is by pulling air with a fan at its exhaust rather than shoving air into its intake.  Thus, the air is being stretched (accelerated) rather than squashed (decelerated).

I see, that makes sense, thanks! You really have to watch out where you read info, because I got this whole idea from here.

@CamilleGoudeseune yes, and maybe even pulling it out the sides (of the wind tunnel) through little holes too

@RobertDiGiovanni, neat!  I found porosity for reducing resonances, but not powered flow through the holes "pulling the air out."  https://www.osti.gov/servlets/purl/1184574  Another question?

@CamilleGoudeseune Laminar windtunnels have a very wide cross section and a throat ahead of the test section so air will accelerate and be laminar around the model. And yes, open tunnels will suck in air with a fan at the aft end of the tunnel.

Well, airflow is relevant to surrounding air so ... wait, what is turbulence?

disturbed flow in a moving stream of air, manifested by variations in speed an direction

Does accelerating air have more turbulence.  One would think not.  It is harder to turn when accelerating.
Can accelerating air cause more turbulence in the surrounding air?
In the aviation sense, air drawn into the accelerated flow will also be accelerated.
Pressure differential is required to accelerate air.  Once the pressure differential "pump" (the wing) is past, there will be turbulence as the entire downwash interacts with even more air and friction inevitably dissipates the energy of the moving air.
Vorticies can last a considerably long time.

does being in a confined space have any effect on this?

yes, because the "confined" space (if it is rigid like a tube) will interact with the moving stream, causing turbulence near the interface as friction slows the fluid stream.
This is spot on with observations of what happens with accelerated air flow over a wing.  The boundary layer will become more and more turbulent towards the trailing edge, but the upper portion of the accelerated airstream is only interacting with air.
As a result, the slowing and increased turbulence of the boundary layer eventually causes the low pressure area over the wing to collapse if Angle of Attack exceeds a critical value.
Interestingly, as seen with helicopter rotor wash (acceleration), the upper boundary of wing accelerated airflow also draws air (momentum) from above, contributing to down wash.


text:
Why is only part of the jet ‘compacted’ when Mach diamonds form?
When Mach diamonds are formed, (in the over expanded case) the flow is firstly compacted by the atmosphere when it leaves the nozzle. This makes a cone-like shape, seen where the flow first exits the nozzle. Picture:

Now my confusion is here: Why is the outlined green part in the picture below not also compacted? What is different about that section of flow?

In other words, why is not all of the flow getting compacted into a shock diamond?

You have drawn a triangle around a flow zone that is downstream of a shock.
The flow in the triangle has two properties

It is parallel to the flow outside the jet flow at that point
The pressure is equal to the ambient

I.e. the boundary between the triangle and the surrounding non-exhaust flow is a slip line.
The flow from the engine exits going straight.  It goes through the shock to enter your triangle.  That shock jumps the pressure to match ambient.  That shock also turns the flow.
Nothing is unusual here for an under expanded nozzle flow.

That is not something that I considered, thanks. Would this situation be much different with over-expanded flow? I incorrectly assumed the pictures were of an overexpanded case.

under-expanded flow starts with a shock and necks in.  over-expanded flow starts with an expansion fan and spreads out.  perfectly expanded flow has no shock/fan features and stays straight.

Ah okay. This other answer must've gotten that mixed up. It says over-expanded flow starts with a shock, and under-expanded starts with a expansion fan. (See the 2 pictures at the bottom).

Maybe I have it backwards....  Lets think through it...  The shock will increase the pressure.  If you have over-expanded, then you've dropped the pressure below ambient -- and it will need a shock to get back up to ambient.  Oops, looks like I got it mixed up.  Everything else I said was right though.


text:
Why will the flow turning angle be 0 (or close to) inside of the Mach cone?
(Disclaimer : This probably seems like my 56th question on the same topic, but this will be my last relating to Mach cone angle and shock angles, as I think I understand most of the topic)
So the reason a shock can never be at less of an angle than the Mach cone is because if it was the flow turning angle would be 0. If it's 0, then there is no shockwave. This makes sense, but this question is asking why the flow turning angle is 0 below the Mach cone angle. This answer provides some insight. Also, here's the graph showing shock angles at different deflection angles.

Also here's a picture I sketched up to illustrate my question:
(Obviously the yellow line would never be a shock with a turning angle of 0)
Why would that yellow line (if it was a shock) have a turning angle of 0 (or a turning angle so low that the shock didn't exist)? It would still have an angle relative to the plane, so it seems like it would still turn flow, allowing a shock to exist. For it to have a turning angle of 0, my intuition tells me the line would have to be parallel to the flow direction.
The answer linked above (here) gives some information about this subject, and it uses the graph above to do that. It explained that the shock angle will never be below the Mach cone angle, but I couldn't seem to find why the flow turning angle will be 0 below the Mach cone angle.

This situation would never occur normally, as the shock angle is always greater than the Mach cone angle, but assume the shock was somehow artificially bent to the angle pictured in the illustration.
Here's a different way of asking this question using the linked answer provided.

an infinitesimally weak shock is at the Mach angle. It does not turn the flow. Any finite strength shock will be at a greater angle and will turn the flow.

Why would the weakest possible shock be right above the Mach cone angle, and the point where the shock disappears be right at the Mach cone angle?

Graph source
Edit : In case there is some confusion, I know the flow always turns near the object because it has to, but this is talking about the flow turning further away from the object.

Why would that yellow line (if it was a shock) have a turning angle of
0 (or a turning angle so low that the shock didn't exist)?

There would be no shock.
There is no shock because there is no turning.  The turning is determined by the obstacle.  A zero degree obstacle has been shown to cause the Mach angle.  More obstacle, more shock angle.
What obstacle is less than a zero obstacle?

Why would the weakest possible shock be right above the Mach cone
angle, and the point where the shock disappears be right at the Mach
cone angle?

Without a disturbance to cause the flow to turn, there is no shock.
Zero disturbance, zero shock.
The object, wall, thing in the flow is the disturbance.  The flow must turn to go around the disturbance.  The flow turning causes the shock.
No wall, no turning, no shock.
It is not the other way around...


text:
Why does the R-27 have reverse taper?

Pictured here is a R-27 missile. If you take a look at the forward fins, they have a reverse taper (meaning they have a higher chord as you extend span).
Aerodynamically speaking, this is the opposite of what you would want for efficiency (I think) so why are they configured this way?

The best information I found claimed that it was for a reduction in spanwise flow. (However I wasn't able to confirm if this info was accurate or not, and it was also using the XF-91 as an example)
My intuition also tells me this would increase the size/strength of the tip vortices, causing more issues, but perhaps also reducing the possibility of tip flow separation.

In terms of efficiency I think the reverse sweep is more efficient, but only slightly. R27's design seems to be based on mechanical structure, not aero efficiency. Remember this is a control surface, and the root is attached to a shaft on a single point. So the "fan-out" style is natural (imagine how you would design the skeleton inside the fins). Fixed delta wings are totally different, where the entire root chord is connected to the body, so the "taper out" style makes sense.

Interesting, what would make the reverse sweep more efficient? Larger tip vortices will induce more drag.

The pic above is from this link and seems to depict a modified R-27, with the fins in question now being middle fins. It can clearly be seen that they are mounted on swivelling axes, and are therefore active control fins, not passive stabilisation fins.
And for active control, we are more considered with an aerodynamic moment about the longitudinal missile axis, than we are with total drag. Therefore, more area away from the centre - creates more control.

And specifically, roll. The fins are to provide maximum rolling torque for the surface area.

It's not obvious to me why roll would matter, as it appears symmetrical about the roll axis.  I would guess that under the hood there's something like ground-imaging radar or something that cares about where it's pointing, but I'm only guessing.  A few words on the subject in the answer would be appreciated, by me at least, is all I'm saying.

@Roger Apparently the thing is able to spin in flight for one reason or other.

Further reading to Koyovis' answer: roll control by differential canard is very difficult

The fact that the front position of the rudders their deflection generates a bevel of air flow affecting the wings installed in the tail in such a way that a certain combination of angles of deflection of rudders, angles of attack and sliding, the phenomenon of reverse control on the roll - the moment from the aerodynamic forces on the wings acts in the direction opposite to the moment from the forces on the steering wheel, and exceeds it. Therefore, on almost all missiles made by the scheme of "duck", the rudders serve only to control the pitch and yaw, and on the channel of the roll are used either stabilizing ailerons or rollers, limiting the speed of rotation of the rocket on the roll.

Vimpel's designers managed to provide control of the rocket through all channels with differentiated deflection of the rudders, abandoning ailerons. For this purpose the K-27 used uniquely shaped butterfly wheels.

That quote doesn't make much sense. What's a bevel of air flow? And what's the scheme of duck? Butterfly wheels?

"duck" = "canard"?


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Why do shocks travel further down the wing as the Mach number increases?
Inspired by this question, why do shocks move further back on a wing as the Mach number increases?
Normal shocks on the surfaces of wings form when the air molecules of the pressure recovery area on the wing don’t have time to react to the incoming air. Not having time to react causes a rapid compression of the air, making a shockwave.
Now if the air on the wing was traveling faster, when it got to the pressure recovery, it would just collapse into a stronger shock but in the same position, right? Why does it not do that and instead move further back in the wing?

Some pictures to see what I’m taking about :

This is when the shock is half down the wing. (Ignore the separation part of the photo)

Then at a higher speed, the shocks for some reason move backwards instead of just staying where they were and getting stronger. Why is that?

@sophit good point. Separation will still cause drag, but not wave drag in this case.

The boundary layer separation aft of the shock wave does not cause wave drag as suggested in the first picture. Wave drag is an intrinsic drag generated at supersonic speeds and has nothing to do with boundary layer. As already suggested by @RobMcDonald beware of random pictures from the internet. Btw, what's the source of that picture?

@sophit The latest response ever, but the source is here

Again...

Why would it stay in the same place?  Is there some physical object to keep it there?  Something like a ridge or a bump or a throat of a nozzle?
On a smooth surface, the location a shock 'chooses' is determined by a balance of effects.  One of those is the upstream Mach number -- as the airfoil goes faster, the supersonic region around the airfoil gets larger and the Mach number in it gets larger.  Another is the back pressure.
Without worrying about the details, when we're faced with a balance of factors like this, there is no reason to think they will be the same as you vary the important conditions.  In fact, it makes a lot more sense for them to be different.
Edit
Adding an image to illustrate the behavior the OP was asking about

I see, but I assumed the back pressure stays the same as the Mach number increases. Is this true? If so, it seems like it’d collapse in the same place?

It would make sense if the back pressure decreased as you speed up, then the shock would travel backwards. Not sure why it would decrease or change though.

It doesn't actually matter if the back pressure stays the same or not.  When the Mach number is higher, the normal shock will be stronger.  A stronger shock means a larger pressure jump at the shock.  Check out the normal shock tables.  https://en.wikipedia.org/wiki/Normal_shock_tables

I came back to this answer because I am a little confused. Imagine you’re going Mach 0.82 like in the picture at the bottom of your answer. When the air accelerates to supersonic, it is forced to decelerate through a normal shock when it arrives at the pressure recovery area. When you increase speed, why wouldn’t it just force this normal shock to be stronger and stay in the same spot?

@Wyatt  Unfortunately, the picture I originally posted does not show the whole picture.  The experimental image was too zoomed in.  Check out the illustration I've now added.  Notice that when you get to supersonic flow, a bow shock forms before the rounded airfoil leading edge.  The bow shock is a normal shock (strong shock solution, subsonic flow downstream) that then bends around and becomes an oblique shock away from the body.  This illustration gives a good indication of the regions of subsonic and supersonic flow.

I see. So if you’re subsonic, increasing speed from Mach .70 -> .90 the shockwave will travel down the wing like in the new picture you added. In this speed range there is no bow shock. In this case, why wouldn’t the shock just increase strength but stay in the same spot relative to the wing?  Sorry if I’m not picking up on something you’ve already answered.

Why would it stay the same strength?  The upstream conditions are changing.

I do think it would change strength. (Perhaps I worded my comment poorly), but I do not see why it would change position on the wing. If the supersonic air is going to collapse into a normal shock, why would it collapse in a different position based on speed? Maybe I should ask a separate question if this is too much to answer in a comments section.

Think of it this way.  The position of the shock is a delicate equilibrium position.  It is like balancing on top of a ball.  Change anything and you will find a different equilibrium position.  The incoming flow (Mach, Pressure) goes through a shock to a higher pressure.  That pressure must match the downstream flow.  The shock will move to find the point where this is true.

I see. So could one way to think of it be that when the lift distribution of the airfoil changes, that in turn changes the shock position? This question has a good picture of what I mean. Basically when those arrows are in different proportions relative to the others, that will change the shock position. Is that correct?

You can't boil most things in fluid dynamics down to a simple cause and effect -- step 1, step 2, step3.  Typically, you must simultaneously solve for 1,2,3 as a coupled problem.  Most people would say the shock causes the pressure distribution around the airfoil, however, the position of the shock does depend on the acceleration (pressure and Mach number) ahead of it.

I see. I think I understand this more now. Definitely agree that a lot of things in fluid mechanics are not simple cause and effect relationships. Thanks.


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Can a shock in certain circumstances be at a lesser angle than the Mach cone?

(Excuse the drawing quality)
What will happen if a shockwave has an angle less than that of the Mach cone?
In this picture, the black line represents the Mach cone. The orange line represents the shockwave angle that would be if it wasn't for the Mach cone making it be the same angle as the Mach cone itself. The yellow line is what this question is about.
So the shockwave can never be in front of the black line (like the orange one is). That's already established, here is an answer that explains it very well.
On a normal occasion, the shockwave would never be at the angle of the yellow line. Best stated from the answer linked above

The shock wave is therefore more "vertical" than the Mach cone

and that is always the case.

So now say that there is a shockwave at the angle of the black line and somehow it got bent past the angle of the Mach cone, just like the yellow line.

Picture of shockwave bending like linked answer above talks about.
If it bent the other way, I would understand how it would stay at the same angle as the Mach cone. But in this case, I don't see anything forcing the shock to be at the same angle as the Mach cone. To summarize, it makes sense why the shock wouldn't be at the angle of the orange line, but I don't see anything stopping it from being at the angle of the yellow line.

By "Certain Circumstances" I meant when the low pressure behind the object will bend the shockwave.
If I can improve this question in any way, please leave a comment. On that note, if the downvote was because this answer to my other question is too similar, I would understand but I don’t think it answers this specific question.

Downvoter, what here did I not follow correctly?

No.
Below is an oblique (2D, Wedge) shock chart.  It shows everything a 2D shock can do.
The top half of the chart is for strong shocks, the bottom half is for weak shocks.  Let's just deal with weak shocks for now.
Consider one value of $M_1$.  Say the $M_1=2.0$ line.  Notice where this line hits the Y-axis.  30 degrees.  For a deflection angle (x axis) of zero degrees.
This is the value of the Mach angle for $M=2.0$.  So, an infinitesimal disturbance that does not turn the flow at all will make a 30 degree shock with a $M=2$ upstream -- this is the definition of the mach cone.
Any flow turning (deflection angle non-zero) and the shock wave angle increases.

Here is the same sort of chart for a conical shock.
The values in this chart are different, but the general shape is the same.  Importantly, the place where a given Mach number intersects the axis is the same -- the Mach angle.
The differences in these charts are due to the 3D relieving effect of a cone.

You should look into NACA Report 1135 -- it is a great source for many compressible flow things.

I see. I understand that the shockwave angle will decrease with lower deflection angles (as stated in the graph), but since the Mach cone is purely a geometric entity, nothing could really change the angle of it besides speed. If the Mach cone stays constant, why does the shockwave not 'misalign' with the angles of it? Actually here is something I just thought of : Are you basically saying because there has to be a non 0 flow turning angle for a shock, and that the shock angle will always have to be greater than the Mach cone angle because of that?

Yes (to the last part).  Or, you could think of it this way -- an infinitesimally weak shock is at the Mach angle.  It does not turn the flow.  Any finite strength shock will be at a greater angle and will turn the flow.

Sorry to comment so long after you've answered, but I was curious about one thing: if the weakest possible shock is at the Mach cone angle, then wouldn't that mean a shock any stronger would have to be at a greater angle than the Mach cone? If so, wouldn’t that violate the rule that the shock angle always has to equal the Mach cone angle?

@Wyatt There is no rule that the shock angle equals the Mach cone angle.  You made that up.  Now that you've revisited an old question, perhaps you should select an answer.  You really should consider going back through all your questions and selecting answers -- some would consider it rude to keep asking questions and never identifying answers.  You aren't holding up your part of the Stack Exchange social contract.

To address the first part of your comment: I'm fairly confident of 2 things: 1. You can't hear an aircraft before the shockwave passes (If the shockwave doesn't travel forwards relative to the plane, neither does the sound) . 2. You can't hear the shockwave, a period of silence, and then the aircraft noise. If those are true, what other angle could the shockwave possibly be at than the Mach cone angle? (The link above is to this answer I just realized)

Second part: I do agree that I should mark answers as accepted, but sometimes I don't remember to click the little check mark. Now that you mention it, I will go through and mark the best answer as accepted. (and also try to make a habit of marking accepted answers) Also, the reason I just keep asking questions on this subject is because for some reason I can't seem to grasp this exact part.

I might just end up writing a question explaining everything I know / think about this subject so you and other people can see more clear where I am at. Typically my questions about shockwaves get downvoted a lot, so not sure if I should. (probably because I've asked quite a few shockwave questions, and people are tired of them)

I think you're just wrapping yourself around the axle...  The shockwave travels at the speed of the aircraft.  Say Mach 2.  The aircraft is supersonic, so no sound comes before the shock.  If you are an observer on the ground, you will hear the shock sometime after the aircraft has passed.   If the aircraft deflects the air 10 deg, the shock will be at 40 degrees.

When you talk about other aircraft noise, I presume you're talking about engine noise, or perhaps a loudspeaker on the airplane that is making a repeated beeping noise.  The sound from that beeper will travel at the speed of sound.  When combined with the forward speed of the aircraft, it will propagate at the Mach angle, in this case 30 degrees.  You will hear the boom first, and the beeping some time later.

Ah that would make sense. I was thinking that the shockwave itself always had to equal the Mach cone angle, but it doesn’t and that makes sense. The oblique shock chart kind of tells you that also. Thanks for helping me figure this out; it’s been confusing for quite some time. (Also, I will be going to mark accepted answers for all of my questions soon)


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Does the forward oblique shock on a supersonic wedge airfoil reduce the lift created by the lower surface?
In the supersonic region of flight, oblique shockwaves form. For this question, I am specifically focusing on the effects of the forward oblique shock (A) as seen in the image below.
My main question revolves around the behavior of airflow on the lower surface of the wedge:
Since the front oblique shock reorients the airflow so that it becomes parallel to the lower (and upper) surface of the wing, does this mean that the lower surface contributes less to lift production at supersonic speeds? Does the flow turning caused by the shockwave reduce the upward force imparted by the lower surface, effectively diminishing its role in overall lift production?

Essentially, because the airflow is made parallel to the lower surface  by the forward oblique shock, does this decrease the amount of air redirected by the lower surface therefore decreasing lift?

Flow is always parallel to a surface.  Otherwise, it would go through the surface -- and that can't happen.

@RobMcDonald Yes, but work has to be done to make the flow parallel to the surface. In subsonic speeds, the work is done by the surface of the wing, and in supersonic speeds the work is done by the shockwave. (Some of the above might be incorrect or inaccurate, as I'm not 100% sure)

The lower surface of a wing is no different from the upper surface.  Aerodynamics doesn't care.  Your question does not include enough information.  Are you considering a blunt nose airfoil with a bow shock out front -- or are you considering a supersonic diamond or wedge airfoil?  Are you talking about a normal shock on the airfoil's curved surface?  BTW, don't forget that flow behind a bow shock is subsonic.

@RobMcDonald Good point. I somehow forgot the bow shock is a normal shock in most cases. Because of that, it doesn't really make sense to use an airfoil in my question so I changed it to a wedge airfoil instead, where the bow shock is more likely to be an oblique shock. Thanks.

You should edit your question further to remove the words about bow and replace with oblique, etc.  Overall, your question is not very clear.  The image you selected is very pretty, but was clearly re-drawn by someone who does not understand superosnic aerodynamics -- the front top surface is drawn essentially parallel to the freestream flow -- which would cause there to be no shock on the top front surface.
Lift is always produced by a pressure difference -- delta P.  In this case of supersonic flow, the highest pressure (by a lot) will be on the lower surface front.

@RobMcDonald I made a few edits. Is it better? I am still planning to edit some more clarity into it.

You've switched to a diagram at zero angle of attack -- where there is no lift at all.  Bottom line 1) the lower surface has no inherent lifting ability.  Lift is caused by a delta in pressure, you can't consider one part in isolation.  2) Supersonic flow has almost nothing to do with subsonic flow.  Everything changes.  In a subsonic flow, the lower surface might see approximately ambient pressure, with acceleration over the top surface causing lower pressure -- delta P causing lift.

In the supersonic flow, the oblique shocks on the top and bottom both cause pressure to increase.  Typically, the bottom surface turns more than the top, so its shock is stronger, and the pressure jump is larger.  DeltaP -- it makes lift.

Does this answer help?

@RobMcDonald Ah okay. Because this question got downvoted, I changed the picture again. This one is actually at AoA so it is making lift, and I added a small summarization at the end. Besides that, I'm having a hard time understanding what else needs improvement. - I did not think of the fact that shockwaves have a pressure jump themselves, which definitely makes sense.

All the words about transitioning from subsonic to supersonic -- and acceleration of the flow -- do not add to the question.  This is a simple question about supersonic flow.  All the other words are distracting fluff.

You can't forget about the pressure jump through a shock -- that is the main thing a shock does.

@RobMcDonald I will make another edit to remove the unnecessary wording. I also think a better response from me would've been "I did not think of the fact that the pressure jump from shockwaves could increase / change the lift". Just bad wording on my part haha.

Since the front oblique shock reorients the airflow so that it becomes parallel to the surface of the wing, does this mean that the lower surface contributes less to lift?

Yes. Since pressure acts perpendicularly to the surface if that surface points for example upward and forward then it produces drag other than lift.
Regarding the diamond airfoil in your picture: Behind a shock wave pressure rises while behind an expansion fan it decreases. That implies that:

on the surfaces B, the aerodynamic force mainly pushes downward and backward;
on the surfaces B', the aerodynamic force mainly pushes upward and backward:
on the surfaces D, the aerodynamic force mainly pulls upward and backward:
and on the surfaces D', the aerodynamic force mainly pulls downward and backward.

The sum of the forces acting backward obviously gives rise to a drag which is the well know "wave drag" typical of supersonic speeds. The sum of the forces acting upward/downward obviously gives the lift.


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What effect does a sideslip have on the lift of a plane?
What effect does a sideslip have on the lift of a plane?

When doing some research prior to this question I found this answer, but it didn't really describe what happens to the wing itself.
Is there some sort of spanwise flow that occurs?

If given thee choices:  1.) Increase lift, 2.) Decrease lift, 3.)  No change - Which one would you instinctively choose?

@MichaelHall I would definitely say 2. But I don’t know why this is the case.

I would agree.  I will let the math experts confirm our intuitions with formulas if that is what you are looking for.  The short answer is along the lines of effectively reducing the span.

@MichaelHall I think I should also definitely rephrase this question, as my main question is about what happens to the wingtips + tip vortices.

Then you should also clarify what you mean by "what happens"...  Because it's equally intuitive that they won't stream straight back perpendicular to the wing, right?

@MichaelHall Yeah good point. I will make another smaller edit to clarify some more.

@Wyatt -- re " as my main question is about what happens to the wingtips + tip vortices."-- among other things, there is a roll torque generated by the displaced tip vortices, even on an unswept wing with zero dihedral with (hypothetically) zero fuselage-wing interference.  Can't remember which direction it is off the top of my head.

@quietflyer Are you talking about the downwash created from the tip vortices? I thought that the tip vortices created downwash, but then I asked Mr.Kämpf and turns out that isn’t accurate. That’s actually what sparked this question.

During a sideslip, the airflow is not parallel to the wing section, with the working aerofoil being altered, and the flow is besides perturbed by the fuselage. The effective wing area decreases, resulting in lower lift and higher drag.


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Which side of an obliquely swept wing will create more lift, and why?
Which side of an obliquely swept wing will create more lift? By "obliquely swept wing" I mean this:

This answer states that the overall lift will decrease due to an effective decrease in wing area. That answer was focusing on an aircraft in a sideslip, but I'm fairly sure the overall effect will be the same as in this picture.

I read through some of this paper, and in section 2.2.2 (AKA page 16) it says:

A well-known effect of wing sweep is the variation of induced downwash along the span from the trailing
wake that produces an additional lift distribution characterized by increased loading on the aft wing and
reduced additional lift on the forward wing. For a wing with no twist or bend, this results in a significant
rolling moment, tending to roll the forward wing downward.

Note: The use of the term "wing sweep" in this excerpt is being used to describe a scenario like the picture above. Not what you would usually think of wing sweep as.
It states that the forward wing will tend to roll downward (so the right wing in the picture above would roll downwards). Why is this?
I can't see anything that would differentiate the aft wing vs. the forward wing in terms of aerodynamics, so what am I missing?

This paper provided by @mins explains a lot, but I am not 100% sure on this. If this was considering a tapered wing like the one above, it would make total sense that the 2 sides have different lift.

Compare swept to unswept, and unswept to forward-swept.  Those affect much more than just lift, though, and their differences vary with speed and AoA, so your answer may turn out to be messy.

This paper helps understanding the rolling moment due to the difference in lift is one among many other moments and side forces.

@mins Thanks for that paper, it almost completely answered my question. So, it says the spanwise flow is the reason for the asymmetrical lift distribution. How does the flow traveling spanwise change the lift? I do have a theory, but not sure about it.

The spanwise component is not accelerated, it creates no lift (but the article focuses on where is the lift concentrated, i.e. near the leading edge and the tip for the left wing). You may be interested in this book which is the story of the oblique wing.

@mins Ah okay, thanks. Also, in the same paper it describes that there is an increase in wave drag on the upwind wing (at supersonic speeds of course). Both sides of the oblique wing are symmetrical, so do you have any idea why this is the case?

Supersonic domain is particular, I've no idea. The shockwaves created by the nose and by the wing tips have likely fixed cones. Each half wing is exposed to these cones differently and the (shock)wave drag may vary.

Does this answer your question?


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What causes this vortex pattern on the slip line of a shockwave?

What are these little vortex patterns from on the slip line of these shocks? At first I thought it was vortex shedding, but it looks a tiny bit different to me (than vortex shedding).
You can see the slip line bending towards the middle (slightly), so would that have something to do with the creation of the vortices? Thanks. (This picture was at Mach 3)
Link to original video.

This flow pattern is typical for shear layers. The flow speed jumps across the layer and mixing causes those regular eddies to form. See here for more.

Graphical representation of a shear layer simulation (source)

Is this a viscous phenomenon?

@sophit good question; I would think it is, not sure.

Everything to do with a shear layer is  viscous phenomena.  This is a transition -- the shear layer is at a Reynolds number where it starts laminar (to the left) but some eddies are forming -- but it isn't quite enough to go fully turbulent.

@RobMcDonald: I was almost certain but then how does this match with the claim from the owner of the simulation claiming to have used "compressible Euler equations (meaning no viscosity)"?

I can't go into depth (it has been more than 20 years since I took a CFD course), but there is a numerical phenomena called 'Artificial Viscosity' where the error in the way the equations are solved can manifest itself in the same way a viscous term would appear in the equations.  Sometimes this is important -- it is what causes an Euler code to separate at the trailing edge of an airfoil without an explicit Kutta condition.  Choosing a method that is either even or odd (order of dominant error term) will result in a method that has either dissipation and dispersion characteristics.


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How is crew resource management trained?
Pilots are taught that pilot monitoring should speak up if something seems wrong, and take over the controls if the pilot flying does not respond adequately, even if the pilot flying is a senior captain. However, do they actually train such situations?
For example recently a final report for accident of US-Bangla Airlines flight 211 was published. The accident involved a badly disoriented captain trying to salvage a really botched approach, and while the rookie copilot did try to be helpful, neither she nor the tower controller was assertive enough to explicitly call go-around.
I would think that some practical training would help in this kind of simulation. I am thinking of a simulator session that would start with the instructor being pilot flying, doing something wrong and acting confused and expecting the trainee to intervene appropriately. The trainee would know the instructor expects them to intervene, and thus wouldn't have qualms about it, but the trained reaction would still help overcome interpersonal issues in real flight.

The positive outcome of "The Miracle on the Hudson" was credited among other things to the good CRM in the cockpit. And it was mentioned that Captain Sullenberger was actually one of the pilots who originally introduced CRM training into their airline and that he was involved in creating and implementing CRM training programs and also regularly taught the course he helped develop. So, CRM training definitely does exist, although none of the articles went into further details as to the specific form of this training.

CRM training which I have been exposed to consists of lecture/videos to cover principles and expand on them with techniques.  There may be some classroom exercises.  However, due to the way that most training is conducted, the "classroom" is usually small numbers.  The application of CRM is in the simulator (SIM), where evaluation of techniques can be provided as the techniques are applied in routine and not-so-routine training scenarios.  Debrief is normally instructor(s) on team, after the SIM session(s).
For the training I get, CRM is pretty effectively integrated into recurrent SIM training.
For ab inito and primary training, the CRM training is similar, with classroom/video training.  Flight schools and academies will have larger class sizes and normally have quite a bit of depth in their training. I recently reviewed a 4 credit hour university course on CRM.
A friend who is a pilot, and also an Emergency Medicine Physician, (and with graduate and undergraduate degrees in Human Factors and Industrial Engineering) has been applying CRM to the hospital emergency department operations, and the military has similar efforts.
CRM has been around for a long time, although there were cultural impediments to it's widespread integration in the aviation context.  Much of that continues to change over the last 20 years.  I expect that CRM will expand into areas such as varying personality styles and other team techniques in the future. An example is some of the people style recognition and skills as developed by Robert Bolton, PhD and Dorthy Grover Bolton.
To summarize, the method of teaching is dependent upon the venue.  In many areas of aviation, CRM is integrated into recurrent training and critiqued on observed operations.  In initial training, the introduction of CRM is largely dependent upon the style of instruction, and whether that instruction is individualized or team based.


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Is certification required to conduct runways repairs?
I know that you need a certain certification to fix a plane, but are you required to have a certification to fix a runway?  Or can anyone with the proper equipment be paid to do it?

To the best of my knowledge, you do not require any certification to fix a runway/taxiway, and anyone with a proper equipment can do that. But to facilitate the process, the FAA has published a few checklists which have to be filled while doing this repair job. These checklists can be found Here. FAA also provides assistance in guiding the construction jobs and identifying risks, the link to which is in the same website as mentioned above.

Indeed no formal certification, but with the amount of problems I have the feeling you could run into with that checklist I would imagine the companies willing to take on this type of job would probably have experience with it. It seems pretty well defined.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/